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Charged circular arc

  1. Jan 22, 2006 #1
    A uniformly charged circular arc AB is of radius R covers a quarter of a circle and is located in the second quadrant. The total charge on the arc is Q > 0. This problem has 4 parts, I got the first 2.
    1. The direction of the electric field E due to the charge distribution at the origin is in quadrant 4.
    2. Determine [tex] \Delta E_x [/tex], the x-component of the electric field vector at the origin O due to the charge element [tex] \Delta q [/tex] locate at an angle [tex] \theta [/tex] subtended by an angular interval [tex]\theta [/tex].
    [tex] \Delta E_x = kQ/R^2 * 2\Delta \theta / \pi * cos \theta [/tex]
    3. Find E_x, the electric field at the origin due to the full arc length for the case where Q= 2.3 [tex]\mu C[/tex] and R= 0.37 m. Answer in units of N/C.

    I have no idea how to find the value for theta. Can someone tell me what I should do?
    Last edited: Jan 22, 2006
  2. jcsd
  3. Jan 23, 2006 #2

    Tom Mattson

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    You're not looking for *a* value of theta, you're looking for a range of values. That's because you have to integrate the expression that you found in part 2.

    So between what two angles is the second quadrant bounded?
  4. Jan 23, 2006 #3
    The arc is bounded between [tex] \pi [/tex] and [tex] \pi /2 [/tex].
    so when I go to integrate it, would it just be the integral just be 2 [tex] \Delta \theta cos \theta [/tex]? Since kQ/r^2 [tex] \pi [/tex] is all constant?
  5. Jan 23, 2006 #4


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    Your latex isn't loading for me...
    cos(theta) is NOT constant, but all the other terms are.
    you have to integrate cos(theta) from 0 to pi/2 .
  6. Jan 23, 2006 #5
    So if I integrate I get
    [tex] KQ/r^2 2 \Delta/ \pi sin \theta [/tex] from pi/2 to 0.
    I'm assuming Delta pi would just be pi/2-0= pi/2.
    So plugging in gives me,
    2.3 x 10^-6 k/(.37)^2 * 2(pi/2)/pi * sin pi/2
    Which = 151205 N/C
    This isn't right.. I think my delta theta is messed up.
    Also, why wouldn't the bounds be pi and pi/2 since it's located in the 2nd quadrant?
  7. Jan 23, 2006 #6


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    What? [tex] \Delta \theta = d \theta [/tex] , which is your integration variable!

    We can't tell where you're measuring theta from , without a diagram.
    The result is the same, either way.
  8. Sep 8, 2009 #7
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